Graphs - College Algebra

Card 0 of 380

Question

Find all horizontal and vertical asymptotes in the graph of $f(x)=\frac{x^2+x-2}{x^2-x-6}$ .

Answer

To find vertical asymptotes, factor each quadratic and then simplify.

$f(x)=\frac{x^2+x-2}{x^2-x-6}=\frac{(x-1)(x+2)}{(x+2)(x-3)}=\frac{x-1}{x-3}$

The vertical asymptotes are the zeros of the denominator, so x=3 is a vertical asymptote.

Horizontal asymptotes are found by analyzing the coefficients of the first term in each equation. The line $y=\frac{a_{n}}{b_{m}}$ (where a is the coefficient of the first term in the numerator and b is the coefficient of the first term in the denominator) is the horizontal asymptote. So we have y=1 as the horizontal asymptote.

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Question

Where are the vertical asymptotes of this rational function?

$g(x) = \frac{x^2+2}{4x^2-8x-5}$

Answer

The vertical asymptotes of a rational function are always the zeroes of the polynomial in the denominator. So to solve this problem all we need to do is find the zeroes of .

We can use the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , for a polynomial . In this case, a=4, b=-8, and c=-5. So we must plug these values into the quadratic formula:

$\frac{-(-8)\pm \sqrt{(-8)^2-4\cdot4\cdot(-5)}}{2\cdot4} = \frac{8\pm \sqrt{64+80}}{8} = \frac{8\pm 12}{8}$

And we get the two roots, .

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Question

Which is a vertical asymptote of the graph of the function

$f(x) = \frac{x^{2}+12x+36}{x^{2}-36}$ ?

(a)

(b)

Answer

The vertical asymptote(s) of the graph of a rational function such as can be found by evaluating the zeroes of the denominator after the rational expression is reduced.

First, factor the numerator and denominator.

The numerator is a perfect square trinomial and can be factored as such:

$x^{2}+12x+36 = x^{2}+2 \cdot x \cdot 6 + 6^{2} = (x+6)^{2} = (x+6)(x+6)$

The denominator can be factored as the difference of squares:

$x^{2}-36 = x^{2}-6 ^{2} = (x+6) (x-6)$

Rewrite

$f(x) = \frac{x^{2}+12x+36}{x^{2}-36}$

as

$f(x) = \frac{ (x+6)(x+6)}{(x+6) (x-6)}$

The expression can be reduced by cancelling in both halves:

$f(x) = \frac{ x+6}{x-6}$

Set the denominator equal to 0 and solve:

The only vertical asymptote is therefore the line of the equation .

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Question

Which is a vertical asymptote of the graph of the function $f(x) = \frac{x-8}{x+ 8}$ ?

(a)

(b)

Answer

The vertical asymptote(s) of the graph of a rational function such as can be found by evaluating the zeroes of the denominator after the rational expression is reduced. The expression is in simplest form, so set the denominator equal to 0 and solve for :

The graph of the line is the only vertical asymptote of the graph of .

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Question

Which of the following is a vertical asymptote of the graph of the function $f(x) = \frac{x^{2}-7x - 18}{x-9}$ ?

(a)

(b)

Answer

The vertical asymptote(s) of the graph of a rational function such as can be found by evaluating the zeroes of the denominator after the rational expression is reduced.

First, factor the numerator. It is a quadratic trinomial with lead term $x^{2}$ , so look to "reverse-FOIL" it as

$x^{2}-7x - 18 = (x- a)(x-b)$

We seek two integers whose sum is and whose product is ; through trial and error, we find and 2, so

$x^{2}-7x - 18 = (x- 9)(x+2)$

Therefore, can be rewritten as

$f(x) = \frac{ (x- 9)(x+2)}{x-9}$

Cancelling , this can be seen to be essentially a polynomial function:

which does not have a vertical asymptote.

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Question

Give the equation(s) of the vertical asymptote(s) of the graph of the function

$f(x) = \frac{x^{2}-3x-10}{x- 2 }$ .

Answer

The vertical asymptote(s) of the graph of a rational function such as can be found by evaluating the zeroes of the denominator.

Set the denominator equal to 0 and solve for :

The only vertical asymptote of the graph is the line of the equation .

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Question

Give the -coordinate of the -intercept the graph of the function

$f(x) = \frac{x^{2}- 5x}{x}$ .

Answer

The -intercept of the graph of a function is the point at which it intersects the -axis. The -coordinate is 0, so the -coordinate can be found by evaluating . This is done by substitution, as follows:

$f(x) = \frac{x^{2}- 5x}{x}$

$f(0) = \frac{0^{2}- 5 \cdot 0}{ 0}$

The value of the denominator here is 0, so is an undefined quantity. Consequently, the graph of has no -intercept.

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Question

Give the -coordinate(s) of the -intercept(s) the graph of the function $f(x) = \frac{x^{2}- 5x}{x}$ .

Answer

The -intercept(s) of the graph of are the point(s) at which it intersects the -axis. The -coordinate of each is 0; their -coordinate(s) are those value(s) of for which , so set up, and solve for , the equation:

$\frac{x^{2}- 5x}{x} = 0$

A fraction is equal to 0 if and only if the numerator is equal to 0, so set

$x^{2} - 5x = 0$

Factor out :

By the Zero Product Property, one of the factors must be equal to 0, so either

or

in which case

.

However, setting in the definition, we see that

$f(0) = \frac{0^{2}- 5 \cdot 0}{0}$ , an undefined expression due to the zero denominator. 5 cannot be eliminated similarly. Therefore, is the only -intercept.

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Question

A function is defined on the domain $\left { 1, 2, 3, 4, 5\right }$ according to the above table.

Define a function . Which of the following values is not in the range of the function $(g \circ f )(x)$ ?

Answer

This is the composition of two functions. By definition, $(g \circ f )(x) = g \left [ f(x)\right ]$ . To find the range of $(g \circ f )(x)$ , we need to find the values of this function for each value in the domain of . Since $(g \circ f )(x) = g \left [ f(x)\right ]$ , this is equivalent to evaluating for each value in the range of , as follows:

Range value: 3

Range value: 5

Range value: 8

Range value: 13

Range value: 21

The range of on the set of range values of - and consequently, the range of $\left (g \circ f \right ) (x)$ - is the set $\left {4 , 10, 19, 34, 58\right }$ . Of the five choices, only 45 does not appear in this set; this is the correct choice.

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Question

Refer to the above diagram, which shows the graph of a function .

True or false: $f\left ( \frac{1}{2} \right ) = 2$ .

Answer

The statement is false. Look for the point on the graph of with -coordinate $\frac{1}{2}$ by going right $\frac{1}{2}$ unit, then moving up and noting the -value, as follows:

$f \left ( \frac{1}{2} \right ) = \frac{3}{2}$ , so the statement is false.

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Question

Define a function $f(x) = \sqrt{4x- 9}$ .

Which statement correctly gives $f^{-1}(x)$ ?

Answer

The inverse function $f^{-1}(x)$ of a function can be found as follows:

Replace with :

$f(x) = \sqrt{4x- 9}$

$y= \sqrt{4x- 9}$

Switch the positions of and :

$x= \sqrt{4y- 9}$

or

$\sqrt{4y- 9} = x$

Solve for . This can be done as follows:

Square both sides:

$\left (\sqrt{4y- 9} \right ) ^{2}= x ^{2}$

$4y- 9 = x ^{2}$

Add 9 to both sides:

$4y- 9+9 = x ^{2}+9$

$4y = x ^{2}+9$

Multiply both sides by $\frac{1}{4}$ , distributing on the right:

$\frac{1}{4} \cdot 4y = \frac{1}{4} \cdot \left (x ^{2}+9 \right )$

$y = \frac{1}{4} \cdot x ^{2}+ \frac{1}{4} \cdot 9$

$y = \frac{1}{4} x ^{2}+ \frac{9}{4}$

Replace with $f^{-1}(x)$ :

$f^{-1}(x)= \frac{1}{4} x ^{2}+ \frac{9}{4}$

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Question

The above diagram shows the graph of function on the coordinate axes. True or false: The -intercept of the graph is

Answer

The -intercept of the graph of a function is the point at which it intersects the -axis (the vertical axis). That point is marked on the diagram below:

The point is about one and three-fourths units above the origin, making the coordinates of the -intercept $\left ( 0, 1\frac{3}{4}\right )$ .

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Question

Evaluate: $\sum_{k =2}^{5} (-1)^{k}(3k+1)$

Answer

Evaluate the expression $(-1)^{k}(3k+1)$ for , then add the four numbers:

$a_{k} = (-1)^{k}(3k+1)$

$a_{2} = (-1)^{2}(3 \cdot 2 +1) = 1 (6+ 1) = 7$

$a_{3} = (-1)^{3}(3 \cdot 3 +1) = -1 (9+ 1) = -10$

$a_{4} = (-1)^{4}(3 \cdot 4 +1) = 1 (12+ 1) = 13$

$a_{5} = (-1)^{5}(3 \cdot 5 +1) =- 1 (15+ 1) = -16$

$\sum_{k =2}^{5} (-1)^{k}(3k+1) = 7 +( -10) + 13 + (-16 ) = -6$

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Question

Evaluate: $\sum_{k =3}^{7} (-2)^{k}$

Answer

Evaluate the expression $\sum_{k =3}^{7} (-2)^{k}$ for , then add the five numbers:

$a_{n} = (-2)^{k}$

$a_{3} = (-2)^{3} = -8$

$a_{4} = (-2)^{4 } =16$

$a_{5} = (-2)^{5 } =-32$

$a_{6} = (-2)^{6 } =64$

$a_{7} = (-2)^{7 } =-128$

$\sum_{k =3}^{7} (-2)^{k} = -8 + 16 + (-32)+ 64 +( -128 ) = -88$

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Question

Consider the polynomial

$2x^{5} + 3 x^{2}- 4x + k$ ,

where is a real constant. For to be a zero of this polynomial, what must be?

Answer

By the Factor Theorem, is a zero of a polynomial if and only if . Here, , so evaluate the polynomial, in terms of , for by substituting 2 for :

$2 \cdot 2^{5} + 3 \cdot 2^{2}- 4 \cdot 2 + k$

$= 2 \cdot 32 + 3 \cdot 4 - 4 \cdot 2 + k$

Set this equal to 0:

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Question

$\left \lfloor N \right \rfloor$ refers to the floor of , the greatest integer less than or equal to .

$\left \lceil N \right \rceil$ refers to the ceiling of , the least integer greater than or equal to .

Define $f(x) = \left \lfloor \frac{1}{2}x- \frac{2}{3} \right \rfloor$ and $g(x) = \left \lceil \frac{2}{3}x- \frac{1}{2} \right \rceil$

Which of the following is equal to $(f \circ g ) \left ( \frac{11}{4} \right )$ ?

Answer

$(f \circ g ) \left ( \frac{11}{4} \right ) =f\left ( g \left ( \frac{11}{4} \right ) \right )$ , so, first, evaluate $g \left ( \frac{11}{4} \right )$ by substitution:

$g(x) = \left \lceil \frac{2}{3}x- \frac{1}{2} \right \rceil$

$g \left ( \frac{11}{4} \right ) = \left \lceil \frac{2}{3} \left ( \frac{11}{4} \right ) - \frac{1}{2} \right \rceil$

$= \left \lceil \frac{11}{6} - \frac{1}{2} \right \rceil$

$= \left \lceil \frac{4}{3} \right \rceil$

$(f \circ g ) \left ( \frac{11}{4} \right ) =f\left ( g \left ( \frac{11}{4} \right ) \right ) = f(2)$ , so evaluate by substitution.

$f(x) = \left \lfloor \frac{1}{2}x- \frac{2}{3} \right \rfloor$

$f(2) = \left \lfloor \frac{1}{2} (2)- \frac{2}{3} \right \rfloor$

$= \left \lfloor1 - \frac{2}{3} \right \rfloor$

$= \left \lfloor \frac{1}{3} \right \rfloor$

,

the correct response.

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Question

$\left \lfloor N \right \rfloor$ refers to the floor of , the greatest integer less than or equal to .

$\left \lceil N \right \rceil$ refers to the ceiling of , the least integer greater than or equal to .

Define $f(x) = \left \lceil 2.2x +6.3 \right \rceil$ and $g (x) = \left \lfloor -2.1 x + 9.3 \right \rfloor$ .

Evaluate $(f \circ g )(3.4 )$

Answer

$(f \circ g )(3.4 ) = f(g(3.4))$ , so first, evaluate using substitution:

$g (x) = \left \lfloor -2.1 x + 9.3 \right \rfloor$

$g (3.4) = \left \lfloor -2.1 (3.4) + 9.3 \right \rfloor$

$= \left \lfloor -7.14+ 9.3 \right \rfloor$

$= \left \lfloor 2.16 \right \rfloor$

$(f \circ g )(3.4 ) = f(g(3.4)) = f(2)$ , so evaluate using substitution:

$f(x) = \left \lceil 2.2x +6.3 \right \rceil$

$= \left \lceil 2.2 (2) +6.3 \right \rceil$

$= \left \lceil 4.4 +6.3 \right \rceil$

$= \left \lceil 10.7 \right \rceil$

,

the correct response.

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Question

Find all horizontal and vertical asymptotes in the graph of $f(x)=\frac{x^2+x-2}{x^2-x-6}$ .

Answer

To find vertical asymptotes, factor each quadratic and then simplify.

$f(x)=\frac{x^2+x-2}{x^2-x-6}=\frac{(x-1)(x+2)}{(x+2)(x-3)}=\frac{x-1}{x-3}$

The vertical asymptotes are the zeros of the denominator, so x=3 is a vertical asymptote.

Horizontal asymptotes are found by analyzing the coefficients of the first term in each equation. The line $y=\frac{a_{n}}{b_{m}}$ (where a is the coefficient of the first term in the numerator and b is the coefficient of the first term in the denominator) is the horizontal asymptote. So we have y=1 as the horizontal asymptote.

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Question

Where are the vertical asymptotes of this rational function?

$g(x) = \frac{x^2+2}{4x^2-8x-5}$

Answer

The vertical asymptotes of a rational function are always the zeroes of the polynomial in the denominator. So to solve this problem all we need to do is find the zeroes of .

We can use the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , for a polynomial . In this case, a=4, b=-8, and c=-5. So we must plug these values into the quadratic formula:

$\frac{-(-8)\pm \sqrt{(-8)^2-4\cdot4\cdot(-5)}}{2\cdot4} = \frac{8\pm \sqrt{64+80}}{8} = \frac{8\pm 12}{8}$

And we get the two roots, .

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Question

Which is a vertical asymptote of the graph of the function

$f(x) = \frac{x^{2}+12x+36}{x^{2}-36}$ ?

(a)

(b)

Answer

The vertical asymptote(s) of the graph of a rational function such as can be found by evaluating the zeroes of the denominator after the rational expression is reduced.

First, factor the numerator and denominator.

The numerator is a perfect square trinomial and can be factored as such:

$x^{2}+12x+36 = x^{2}+2 \cdot x \cdot 6 + 6^{2} = (x+6)^{2} = (x+6)(x+6)$

The denominator can be factored as the difference of squares:

$x^{2}-36 = x^{2}-6 ^{2} = (x+6) (x-6)$

Rewrite

$f(x) = \frac{x^{2}+12x+36}{x^{2}-36}$

as

$f(x) = \frac{ (x+6)(x+6)}{(x+6) (x-6)}$

The expression can be reduced by cancelling in both halves:

$f(x) = \frac{ x+6}{x-6}$

Set the denominator equal to 0 and solve:

The only vertical asymptote is therefore the line of the equation .

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